I would like a query using regexp_like within Oracle's SQL which only validates uppercase characters [A-Z] and numbers [0-9]
SELECT *
FROM dual
WHERE REGEXP_LIKE('AAAA1111', '[A-Z, 0-9]')
List item
The select Statement probalby should look like
SELECT 'Yes' as MATCHING
FROM dual
WHERE REGEXP_LIKE ('AAAA1111', '^[A-Z0-9]+$')
Which means that starting from the very first ^ to the last $ letter every character should be upper case or a number. Important: no comma or space between Z and 0. The + stands for at least one or more characters.
Edit: Based on the answer of Barbaros another way of selecting would be possible
SELECT 'Yes' as MATCHING
FROM DUAL
WHERE regexp_like('AAAA1111','^[[:digit:][:upper:]]+$')
Edit: added a DBFiddle
A quick help may be found here and for oracle regular expressions here.
You can use :
select str as "Result String"
from tab
where not regexp_like(str,'[[:lower:] ]')
and regexp_like(str,'[[:alnum:]]')
where not regexp_like with POSIX [^[:lower:]] pattern stands for eliminating the strings
containing lowercase,
and regexp_like with POSIX [[:alnum:]] pattern stands for accepting the strings
without symbols
( containing only letters and numbers even doesn't contain a space because of the trailing space at the end part of [[:lower:] ] )
Demo
Related
I am having following string in my query
.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt
beginning with a period from which I need to extract the segment between the final \ and the file extension period, meaning following expected result
ABC__123_123_123_ABC123
Am fairly new to using REGEXP and couldn't help myself to an elegant (or workable) solution with what Q&A here or else. In all queries the pattern is the same in quantity and order but for my growth of knowledge I'd prefer to not just count and cut.
You can use REGEXP_REPLACE function such as
REGEXP_REPLACE(col,'(.*\\)(.*)\.(.*)','\2')
in order to extract the piece starting from the last slash upto the dot. Preceding slashes in \\ and \. are used as escape characters to distinguish the special characters and our intended \ and . characters.
Demo
You need just regexp_substr and simple regexp ([^\]+)\.[^.]*$
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'([^\]+)\.[^.]*$',
1, -- position
1, -- occurence
null, -- match_parameter
1 -- subexpr
) substring
from dual;
([^\]+)\.[^.]*$ means:
([^\]+) - find one or more(+) any characters except slash([] - set, ^ - negative, ie except) and name it as group \1(subexpression #1)
\. - then simple dot (. is a special character which means any character, so we need to "escape" it using \ which is an escape character)
[^.]* - zero or more any characters except .
$ - end of line
So this regexp means: find a substring which consist from: one or more any characters except slash followed by dot followed by zero or more any characters except dot and it should be in the end of string. And subexpr parameter = 1, says oracle to return first subexpression (ie first matched group in (...))
Other parameters you can find in the doc.
Here is my simple full compatible example with Oracle 11g R2, PCRE2 and some other languages.
Oracle 11g R2 using function substr (Reference documentation)
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}',
1,
1
) substring
from dual;
Pattern: ((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}
Result: ABC__123_123_123_ABC123
Just as simple as it can be, regular expressions always follow a minimal standard, as you can see portability also provided, just for the case someone else is interested in going the simplest way.
Hopefully, this will help you out!
I need to find rows where the phone number field contains unexpected characters.
Most of the values in this field look like:
123456-7890
This is expected. However, we are also seeing character values in this field such as * and #.
I want to find all rows where these unexpected character values exist.
Expected:
Numbers are expected
Hyphen with numbers is expected (hyphen alone is not)
NULL is expected
Empty is expected
Tried this:
WHERE phone_num is not like ' %[0-9,-,' ' ]%
Still getting rows where phone has numbers.
from https://regexr.com/3c53v address you can edit regex to match your needs.
I am going to use example regex for this purpose
select * from Table1
Where NOT REGEXP_LIKE(PhoneNumberColumn, '^[+]*[(]{0,1}[0-9]{1,4}[)]{0,1}[-\s\./0-9]*$')
You can use translate()
...
WHERE translate(Phone_Number,'a1234567890-', 'a') is NOT NULL
This will strip out all valid characters leaving behind the invalid ones. If all the characters are valid, the result would be NULL. This does not validate the format, for that you'd need to use REGEXP_LIKE or something similar.
You can use regexp_like().
...
WHERE regexp_like(phone_num, '[^ 0123456789-]|^-|-$')
[^ 0123456789-] matches any character that is not a space nor a digit nor a hyphen. ^- matches a hyphen at the beginning and -$ on the end of the string. The pipes are "ors" i.e. a|b matches if pattern a matches of if pattern b matches.
Oracle has REGEXP_LIKE for regex compares:
WHERE REGEXP_LIKE(phone_num,'[^0-9''\-]')
If you're unfamiliar with regular expressions, there are plenty of good sites to help you build them. I like this one
In my column, there is a string that contains a word which starts with character '=' How can I extract those words? I found REGEXP_SUBSTR but I couldn't find out particular regular expression to do this? I appreciate any help. Thanks.
EDIT :
I have such a string :
"What a =lovely day!"
I want to get "=lovely"
You can use regexp_substr for this:
select regexp_substr(col, '=\S+')
from your_table;
=\S+:
= - match literal =
\S+ - match one or more non space characters
Try
REGEXP_SUBSTR('What a =lovely day!', '=\w+')
or
REGEXP_SUBSTR('What a =lovely day!', '=\S+')
depending on your needs.
Also if you want to match based on a list of special characters, use something like below.
In this example, you can match = and #. You can add more special character if you like.
Also if you want just = to be returned in case rest word is missing after that, then use \S*. Else use \S+
For strings which dont have this format, you will get null.
select regexp_substr(col1,'[=#]\S*') from
(select 'what a =lovely day' as col1 from dual union all
select 'some other #word as' from dual union all
select 'a normal string' from dual)
Here's the code that is in production:
dynamic_sql := q'[ with cte as
select user_id,
user_name
from user_table
where regexp_like (bizz_buzz,'^[^Z][^Y6]]' || q'[') AND
user_code not in ('A','E','I')
order by 1]';
Start at the beginning and search bizz_buzz
Match any one character that is NOT Z
Match any two characters that are not Y6
What's the ']' after the 6?
Then what?
I think that StackOverflow's formatting is causing some of the confusion in the answers. Oracle has a syntax for a string literal, q'[...]', which means that the ... portion is to be interpreted exactly as-is; so for instance it can include single quotes without having to escape each one individually.
But the code formatting here doesn't understand that syntax, so it is treating each single-quote as a string delimiter, which makes the result look different that how Oracle really sees it.
The expression is concatenating two such string literals together. (I'm not sure why - it looks like it would be possible to write this as a single string literal with no issues.) As pointed out in another answer/comment, the resulting SQL string is actually:
with cte as
select user_id,
user_name
from user_table
where regexp_like (bizz_buzz,'^[^Z][^Y6]') AND
user_code not in ('A','E','I')
order by 1
And also as pointed out in another answer, the [^Y6] portion of the regex matches a single character, not two. So this expression should simply match any string whose first character is not 'Z' and whose second character is neither 'Y' nor '6'.
When not in couples ] means... Well... Itself:
^[^Z][^Y6]]/
^ assert position at start of the string
[^Z] match a single character not present in the list below
Z the literal character Z (case sensitive)
[^Y6] match a single character not present in the list below
Y6 a single character in the list Y6 literally (case sensitive)
] matches the character ] literally
Start at the beginning and search bizz_buzz
Match any one character that is NOT Z
Match any two one characters that is not Y or 6
What's the ']' after the 6? it's a ]
I'm afraid I have to post this here as the comment section is inappropriate for the formatting required. After your edit above that shows the entire statement, I ran this to see what the string ends up being:
select q'[ with cte as
select user_id,
user_name
from user_table
where regexp_like (bizz_buzz,'^[^Z][^Y6]]' || q'[') AND
user_code not in ('A','E','I')
order by 1]' txt
from dual;
It ended up yielding this:
with cte as
select user_id,
user_name
from user_table
where regexp_like (bizz_buzz,'^[^Z][^Y6]') AND
user_code not in ('A','E','I')
order by 1
It is apparent now that the closing bracket and quote at the end of the regex belong to the first alternate quote string and not to the regex. This is concatenating 2 alternate quoted strings which is a tad confusing as it sure looked like part of the regex. If anything you are learning the importance of comments for the poor person behind you! Please comment this accordingly when you are done figuring this out. Even include a link to this post.
I need to only display those strings (name of manufacturers) that contain 2 or more identical vowels in Oracle11g. I am using a RegEx to find this.
SELECT manuf_name "Manufacturer", REGEXP_LIKE(manuf_name,'([aeiou])\2') Counter FROM manufacturer;
For example:
The RegEx accepts
OtterBox
Abca
abcA
The RegEx rejects
Samsung
Apple
I am not sure how to proceed ahead.
I think you want something like this:
WITH mydata AS (
SELECT 'OtterBox' AS manuf_name FROM dual
UNION ALL
SELECT 'Apple' FROM dual
UNION ALL
SELECT 'Samsung' FROM dual
)
SELECT * FROM mydata
WHERE REGEXP_LIKE(manuf_name, '([aeiou]).*\1', 'i');
I am not sure why you used \2 as a backreference instead of \1 -- \2 doesn't refer to anything in this regex. Also, note the wildcard and quantifier .* to indicate that there can be any number of any character between the first occurrence of the vowel and the second. Third, note the 'i' parameter to indicate a case-insensitive search (which I think is what you want since you say that the regex should match "OtterBox").
SQL Fiddle here.
David yours wasn't quite working for me. What about this?
\w*([aeiou])\w*\1+\w*
https://regex101.com/r/eE3iC2/3
EDIT: updated one per suggestions:
.*([aeiou]).*\1.*
https://regex101.com/r/eE3iC2/5